Zeros distribution of gaussian entire functions

arXiv:1401.2776v1 [math.CV] 13 Jan 2014

Zeros distribution of gaussian entire functions A. O. Kuryliak Department of Mechanics and Mathematics, Ivan Franko National University of L’viv, Ukraine [email protected] O. B. Skaskiv Department of Mechanics and Mathematics, Ivan Franko National University of L’viv, Ukraine [email protected] Abstract In this paper we consider a random entire function of the form P+∞ n f (z, ω) = n=0 ξn (ω)an z , where ξn (ω) are independent standard complex gaussian random variables and an ∈ C satisfy the relations p n lim |an | = 0 and #{n : an 6= 0} = +∞. We investigate asymp-

n→+∞

totic properties of the probability P0 (r) = P {ω : f (z, ω) has no zeros inside rD}. p0 (r) = ln− P0 (r), N (r) = #{n : ln(|an |r n ) > 0}, P+∞Denote s(r) = n=0 ln+ (|an |r n ). Assuming that a0 6= 0 we prove that ln(p0 (r) − s(r)) , ln s(r) r→+∞

0 ≤ lim

r ∈E /

lim

r→+∞ r ∈E /

1 ln(p0 (r) − s(r)) ≤ , r→+∞ ln s(r) 2 lim

r ∈E /

ln(p0 (r) − s(r)) = 1. ln N (r)

where E is a set of finite logarithmic measure. Remark that the previous inequalities are sharp. Also we give an answer to open question from [25, p. 119].

Subject Classification: 30B20, 30D35, 30E15 Keywords: gaussian entire functions, hole probability, zeros distribution of random entire functions

1

Introduction

One of the problems on random functions is investigation of value distribution of there functions and also asymptotic properties of the probability of absence

2

A. O. Kuryliak, O. B. Skaskiv

of zeros in a disc (“hole probability”). These problems were considered in papers of J. E. Littlewood and A. C. Offord ([1]–[6]), M. Sodin and B. Tsirelson ([7]–[9]), Yu. Peres and B. Virag ([10], [11]), P. V. Filevych and M. P. Mahola ([12]–[14]), M. Sodin ([15]–[17]), F. Nazarov, M. Sodin and A. Volberg ([18], [19]), M. Krishnapur ([20]), A. Nishry ([21]–[25]) and many others. So, in [9] it was considered a random entire function of the form ψ(z, ω) =

+∞ X

zk ξk (ω) √ , k! k=0

(1)

where {ξk (ω)} are independent complex valued random variables with the density function 1 2 pξk (z) = e−|z| , z ∈ C, k ∈ Z+ . π We denote such a distribution by NC (0, 1). Let us denote by nψ (r, ω) the counting function of zeros of the function ψ(z, ω) in rD = {z : |z| < r}. Then ([9]) for any δ ∈ (0, 1/4] and all r ≥ 1 we have n n(r, ω) o P ω: − 1 ≥ δ ≤ exp(−c(δ)r 4 ), 2 r where the constant c(δ) depends only on δ. Also in [9] it was investigated the probability of absence of zeros of the function ψ(z, ω), P0 (r) = P {ω : ψ(z, ω) 6= 0 inside rD}. In particular, it was proved in [9] that there exist constants c1 , c2 > 0 such that exp(−c1 r 4 ) ≤ P0 (r) ≤ exp(−c2 r 4 ) (r ≥ 1). Also in [9] the authors put the following question: Does there exist the limit ln− P0 (r) ? r→+∞ r4 lim

We find the answer to this question in [21]. For the function ψ(z, ω) it was proved that 3e2 ln− P0 (r) lim = . r→+∞ r4 4 In [21] it was proved that if all of ξn (ω) : ξn (Ω) ⊂ K, where K ⊂ C and 0 6∈ K then there exists r0 (K) < +∞ such that ψ(z, ω) must vanish somewhere in the disc r0 D. For the function of the form (1) one can fix the disc of radius r and ask for the asymptotic behaviour of P {ω : nψ (r, ω) ≥ m} as m → +∞. So in [20] it was proved, that for any r > 0 we get 1 ln P {ω : nψ (r, ω) ≥ m} = − m2 ln m(1 + o(1)) (m → +∞). 2

3

Zeros distribution of gaussian entire functions

Very large deviations of zeros of function (1) were also considered in [19]. There we find such a relation    1  1 2α − 1, 2 ≤ α ≤ 1; ln ln P {ω : |nψ (r,ω)−r2 |>rα} = 3α − 2, 1 ≤ α ≤ 2; lim r→+∞  ln r  2α, α ≥ 2.

More generally, in [24, 25] it was considered gaussian entire functions of the form +∞ X f (z, ω) = ξk (ω)an z n , n=0

p where a0 6= 0, n ∈ Z+ , lim n |an | = 0. If ε > 0, then there exists ([24, 25]) n→+∞ R a set of finite logarithmic measure E ⊂ (1, +∞) ( E drr < +∞) such that for all r ∈ (1, +∞) \ E we obtain s(r) − s

1/2+ε

(r) ≤ p0 (r) ≤ s(r) + s

1/2+ε

(r), s(r) =

+∞ X

ln+ (an r n ).

(2)

n=0

One can find similar results for gaussian analytic functions in the unit disc in [20], [10], [11], [17]. Also in [25, p. 119], it was formulated the following question: Is the error term in inequality (2) optimal for a regular sequence of the coefficients {an }? The aim of this paper is to answer this question.

2

Notation

In this section we consider the functions of the form f (z, ω) =

+∞ X

ξn (ω)an z n ,

(3)

n=0

where ξn (ω)p∈ NC (0, 1) and an ∈ C, n ∈ Z+ such that #{n : an 6= 0} = +∞, lim n |an | = 0. n→+∞

In this paper we study asymptotic behaviour of

p0 (r) = ln− P {ω : nf (r, ω) = 0} as r → +∞ for random entire functions of the form (3).

4

A. O. Kuryliak, O. B. Skaskiv

For r > 0, δ > 0 denote N ′ = {n : an = 0}, N (r) = {n : ln(|an |r n ) > 0}, Nδ (r) = {n : ln(|an |r n ) > −δn}, N(r) = #N (r), Nδ (r) = #Nδ (r), X X X m(r) = n, mδ (r) = n, s(r) = 2 ln(|an |r n ), n∈N (r)

n∈Nδ (r)

n∈N (r)

n

µf (r) = max{|an |r : n ∈ Z+ }, νf (r) = max{n : µf (r) = |an |r n }, Mf (r) = max{|f (z)| : |z| ≤ r}, Sf2 (r) =

3

+∞ X n=0

|an |2 r 2n .

Auxiliary lemmas

Lemma 3.1 (Borel-Nevanlinna, [27]). Let u(r) be a nondecreasing continuous function on [r0 ; +∞) and limr→+∞ u(r) = +∞, and ϕ(u) be a continuous nonincreasing positive function defined on [u0 ; +∞) and 1) u0 = u(r0 ); 2) lim ϕ(u) = 0; u→+∞

3)

R +∞ u0

ϕ(u)du < +∞.

Then for all r ≥ r0 outside a set E of finite measure we have u{r − ϕ(u(r))} > u(r) − 1. We need the following elementary corollary of this lemma. Lemma 3.2. There exists a set E ⊂ (1; +∞) of finite logarithmic measure such that for all r ∈ (1; +∞) \ E we obtain p p m(re−δ ) > em(r) exp{−2 ln m(r)}, m(reδ ) < em(r) exp{2 ln m(r)},

where δ =

1 . 2 ln m(r)

Lemma 3.3. Let ε > 0. There exists a set E ⊂ (1; +∞) of finite logarithmic measure such that for all r ∈ (1; +∞) \ E we have p (4) N(r) < s1/2 (r) exp{(1 + ε) ln s(r)}. Proof. Remark that (see also [24])

N−δ (r) = #{n : |an |r n ≥ eδn } = #{n : |an |(re−δ )n ≥ 1} = N(re−δ )

5

Zeros distribution of gaussian entire functions

and

N (r)−1

m(r) =

X

n∈N (r)

n≥

X

n=

n=0

N 2 (r) (N(r) − 1)N(r) > 2 e

for r > r0 , where N(r0 ) > 4. So we obtain for r ∈ (1; +∞) \ E X s(r) = ln(|an |r n ) ≥ 2 n∈N (r)

Then

X

n∈N−δ (r)

ln(|an |r n ) ≥

X

nδ =

n∈N−δ (r)

p e m(r) exp{−2 ln m(r)}. = δm(re−δ ) > 2 ln m(r) p ln s(r) > 1 + ln m(r) − 2 ln m(r) − ln ln m(r)

and for r > r2 we get ln m(r) < 2 ln s(r). So for any ε > 0 p s(r) > em(r) exp{−2 ln m(r) − ln ln m(r)} > p N 2 (r) exp{−2 (1 + ε) ln s(r) − ln((1 + ε) ln s(r))} > >e e p > N 2 (r) exp{−(2 + 2ε) ln s(r)}, as r → +∞ outside some set of finite logarithmic measure.

Let us note that the exponent 1/2 in the inequality (4) can not be replaced by a smaller number. Lemma 3.4. There exist a random entire function of the form (3) and a set E ⊂ (1; +∞) of finite logarithmic measure such that for all r ∈ (1; +∞) \ E we have p s(r) N(r) > 3 . ln s(r) Proof. Choose f (z) = 1 +

+∞ X zn . n n 2 ( ) n=1 2

Then the function y(n) = ln an = − n2 ln( n2 ) is concave function and the sequence {an } is log-concave([22], [26]). Also by Stirling’s formula we obtain +∞ +∞ +∞ 2m +∞ X X X X rn rn r r 2m Mf (r) = 1 + > 1 + = n > 1 + n = 1 + n 2 n 2 m m m m!e ( ) ( ) 2 2 n=1 m=1 m=1 n/2=1 n r2 o 2 r , ln Mf (r) > . = exp e e

6

A. O. Kuryliak, O. B. Skaskiv

By Wiman-Valiron’s theorem there exists a set E of finite logarithmic measure such that for all r ∈ (1; +∞) \ E we get ln µf (r) + ln ln µf (r) > ln Mf (r) > r 2 /e, ln µf (r) > r 2 /2e and finally  r2 1  r2 < ln µf (r) = ln aν + νf (r) ln r, νf (r) > − ln aν > r, r → +∞. 2e ln r 2e Therefore, outside a set E of finite logarithmic measure we get ([22]) s(r) < 2(N(r) + 1) ln µf (r) < ln2 µf (r)(ln ln µf (r))2 = ln2 µf (r) (ln ln µf (r))2 < ln3 rνf2 (r) ln2 νf (r) < ln r ln2 r < νf2 (r) ln5 νf (r) < N 2 (r) ln5 N(r) < N 2 (r) ln5 s(r), s p s(r) s(r) > 3 . N(r) > 5 ln s(r) ln s(r)

= ln3 r

Also we will use the following lemma. Lemma 3.5. Let {ζn (ω)} be a sequence of independent identically distributed random variables, such that M|ζn | < +∞ and M( |ζ1n | ) < +∞, n ∈ Z+ . Then io n h1 ≤ |ζn (ω)| ≤ n = 1. P ω : (∃N ∗ (ω))(∀n > N ∗ (ω)) n Proof. Let F|ζ| (t) = F|ζn | (t) be the distribution function of the random variable |ζn |, n ∈ Z+ . Denote Bm = {ω : |ζm(w)| ≥ m}, m ∈ Z+ . Then +∞ X

m=1

P {ω : |ζm (w)| ≥ m} = =

s +∞ X X

Z

+∞ Z X

m=1 |t|≥m

dF|ζ| (t) =

s=1 m=1 |t|∈[s,s+1)

≤

+∞ X s=1

dF|ζ| (t) =

Z

+∞ X +∞ X

Z

dF|ζ| (t) =

m=1 s=m |t|∈[s,s+1) +∞ X s=1

s

Z

dF|ζ| (t) ≤

|t|∈[s,s+1)

|t|dF|ζ|(t) = M|ζ| < +∞.

|t|∈[s,s+1)

P Therefore we obtain +∞ m=1 P (Bm ) < +∞. So, by the Borel-Cantelli lemma only finite quantity of the events Bn may occur. Then n h io ∗ ∗ P (A1 ) = P ω : (∃N1 (ω))(∀n > N1 (ω)) |ζn (ω)| ≤ n = 1.

7

Zeros distribution of gaussian entire functions 1 Since M( |ζ| ) < +∞, we get similarly for random variable

1 |ζ(ω)|

io n h 1 ∗ ∗ ≤n = P (A2 ) = P ω : (∃N2 (ω))(∀n > N2 (ω)) |ζn (ω)| n h 1 io = P ω : (∃N2∗ (ω))(∀n > N2∗ (ω)) |ζn (ω)| ≥ = 1. n Finally, io n h1 ≤ |ζn (ω)| ≤ n = 1, P (A1 ∩ A2 ) = P ω : (∃N ∗ (ω))(∀n > N ∗ (ω)) n where N ∗ (ω) = max{N1∗ (ω), N2∗ (ω)}. The random variables ξn ∈ NC (0, 1), n ∈ Z+ satisfy conditions of Lemma 3.5. So, we get the following result. Lemma 3.6. Let ξn ∈ NC (0, 1), n ∈ Z+ . Then n h1 io P ω : (∃N ∗ (ω))(∀n > N ∗ (ω)) ≤ |ξn (ω)| ≤ n = 1. n

4

Upper and lower bounds for p0(r)

Theorem 4.1. Let ε > 0 and f (z, ω) be random entire function of the form (3) with a0 6= 0. There exists a set E ⊂ (1; +∞) of finite logarithmic measure such that for all r ∈ (1; +∞) \ E we have p p0 (r) ≤ s(r) + N(r) exp{(2 + ε) ln N(r)}.

Proof. Similarly as in [24], we consider the event Ω1 = ∩4i=1 Ai , where p n 2eN 1/3 (r) exp{2 ln N(r)} o A1 = ω : |ξ0 (ω)| ≥ , |a0 | io n h 1 , A2 = ω : (∀n ∈ N (r) \ {0}) |ξn (ω)| ≤ |an |r n N 2/3 (r) n h 1 io A3 = ω : (∀n ∈ Nδ (r) \ (N (r) ∪ {0})) |ξn (ω)| ≤ 2/3 , N (r) n h io 1 . A4 = ω : (∀n 6∈ Nδ (r) ∪ N ′ ∪ {0}) |ξn (ω)| ≤ n , δ = 2 ln N(r)

(5)

8

A. O. Kuryliak, O. B. Skaskiv

If Ω1 occurs, then for r 6∈ E we obtain +∞ X p n ξn (ω)an r ≥ 2eN 1/3 (r) exp{2 ln N(r)}− |ξ0 (ω)a0 | − n=1 n X X X |an |r |an |r n − − − ne−nδ > n 2/3 2/3 |an |r N (r) N (r) ′ n∈N (r)

n6∈Nδ (r)∪N

n∈Nδ (r)\N (r)

X p > 2eN 1/3 (r) exp{2 ln N(r)} −

n∈Nδ (r)

Z

1

N 2/3 (r)

−

+∞

p xe−δx dx > 2eN 1/3 (r) exp{2 ln N(r)}− 1 p −N 1/3 (r) − eN 1/3 (r) exp{2 ln N(r)} − 8 ln2 N(r) > 0 −

as r → +∞, because Z

1

+∞ −δx

xe

2 e−δ dx = 2 (δ + 1) < 2 = 8 ln2 N(r). δ δ

So, we proved that first term dominants the sum of all the other terms inside rD, i.e. +∞ X n (6) ξn (ω)an z . |ξ0(ω)a0 | > n=1

If Ω1 occurs then the function f (z, ω) has no zeros inside rD. Now we find a lower bound for the probability of the event Ω1 . n 4e2 N 2/3 (r) exp{4pln N(r)} o , P (A1 ) = exp − |a0 |2 Y Y 1 1 P (A2 ) ≥ = × 2 2n 4/3 2|an | r N (r) 2|an |2 r 2n n∈N (r) n∈N (r) o n 4 × exp{−N(r) ln(N 4/3 (r))} = exp −s(r) − N(r) ln N(r) − N(r) ln 2 , 3 n o Y 1 δ 4/3 ≥ exp −N(re ) ln(2N (r)) ≥ P (A3) ≥ 2N 4/3 (r) δ n∈N (re ) n o p ≥ exp −eN(r) exp{2 N(r)} ln(2N 4/3 (r)) , P (A4) = P {ω : (∀n 6∈ Nδ (r) ∪ N ′ ∪ {0})|ξn (ω)| < n} ≥ X 1 2 ≥1− e−n > , r → +∞. 2 ′ n6∈Nδ (r)∪N ∪{0}

9

Zeros distribution of gaussian entire functions

It follows from Ω1 ⊂ {ω : n(r, ω) = 0} that for any ε > 0 and for every r ∈ [r0 , +∞) \ E p0 (r) = ln− P {ω : n(r, ω) = 0} ≤ ln− P (Ω1 ) =

4 X n=1

ln− P (An ) ≤

p 4e N (r) exp{4 ln N(r)} ≤ ln 2 + + s(r) + 2N(r) ln N(r) + N(r) ln 2+ |a0 |2 p p +eN(r) exp{2 N(r)} ln(2N 4/3 (r)) ≤ s(r) + N(r) exp{(2 + ε) N(r)} 2

2/3

A random entire function of the form g(z, ω) =

+∞ X

e2πiθn (ω) an z n

(7)

n=0

where independent random variables θn (ω) are uniformly distributed on (0, 1), was considered in [14]. For such functions there were proved the following statements. Theorem 4.2 ([12]). Let g(z, ω) be a random entire function of the form (7). Then for r > r0 we obtain Ng (r, ω) ≤ where

1 Ng (r, ω) = 2π

and k = min{n ∈ Z+ : an 6= 0}.

Z

0

1 + ln Sg (r), 2e

2π

ln |g(reiα , ω)|dα − ln |ak |,

Theorem 4.3 ([14]). There exists an absolute constant C > 0 such that for a function g(z, ω) of the form (7) almost surely we have ln Sg (r) ≤ Ng (r, ω) + C ln Ng (r, ω), r0 (ω) ≤ r < +∞.

(8)

Corollary 4.4. Let (ζn (ω)) be a sequence of independent identically distributed random variables such that for any n ∈ N random variable arg ζn (ω) is uniformly distributed on [−π, π) and M|ξn | < +∞, M( |ξ1n | ) < +∞, n ∈ Z+ . Then there exists an absolute P constant C > 0 such that for every random n function of the form f (z, ω) = +∞ n=0 ζn (ω)an z we get almost surely Z 2π 1 ln |f (reiα, ω)|dα − ln |ak ζk (ω)| ≥ ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω) 2π 0 for r0 (ω) ≤ r < +∞ and k = min{n ∈ Z+ : an 6= 0}.

10

A. O. Kuryliak, O. B. Skaskiv

Since random variables arg ξn (ω)(here ξn (ω) ∈ NC (0, 1)) are also uniformly distributed on [−π, π), we have the following statement for the functions of the form (3). Corollary 4.5. There exists an absolute constant C > 0 such that for the functions of the form (3) we get almost surely Z 2π 1 ln |f (reiθ , ω)|dθ − ln |ak ξk (ω)| ≥ ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω) 2π 0 for r0 (ω) ≤ r < +∞ and k = min{n ∈ Z+ : an 6= 0}. Proof of corollary 4.4. It follows from Theorem 4.2 that ln Ng (r, ω) ≤ 1 + ln ln Sg (r) and by Theorem 4.3 we have almost surely Ng (r, ω) ≥ ln Sg (r) − C ln Ng (r, ω) ≥ ln Sg (r) − (C + 1) ln ln Sg (r), for r0 (ω) ≤ r < +∞. Therefore P {ω : (∃r0 (ω))(∀r > r0 (ω)) [Ng (r, ω) ≥ ln Sg (r) − (C + 1) ln ln Sg (r)]} = 1. Consider a random function f (z, ω1 , ω2 ) =

+∞ X

εn (ω1 )ηn (ω2 )an z n ,

n=0

where εn (ω1 ) = eiθn (ω1 ) , θn (ω1 ) and arg ηn (ω2 ) are uniformly distributed on [−π, π). Also both sequences {εn (ω1 )}, {ηn (ω2 )} are sequences of independent random variables defined on the Steinhaus probability spaces (Ω1 , A1 , P1 ) and (Ω2 , A2 , P2 ), respectively. Define Af = {(ω1, ω2 ) : (∃r0 (ω1 , ω2 ))(∀r > r0 (ω1 , ω2 )) [Nf (r, ω1 , ω2 ) ≥ ln Sf (r, ω2 ) − (C + 1) ln ln Sf (r, ω2 )]}, where Sf2 (r, ω2 )

=

+∞ X n=0

2

2

2 2n

|εn (ω1 )| |ηn (ω2 )| |an | r

=

+∞ X n=0

|ηn (ω2 )|2 |an |2 r 2n .

Consider the events n F = {ω2 : (∀n ∈ N) [ηn (ω2 ) 6= 0]}, H = ω2 :

lim

n→+∞

o p n |an ||ηn (ω2 )| = 0 .

Then by Lemma 3.5 P2 (H) = 1. Since M( |ξ1n | ) < +∞, the probability of the event F +∞ X 1 ≥ P2 (F ) ≥ 1 − P2 {ω2 : ηn (ω2 ) = 0} = 1. n=0

11

Zeros distribution of gaussian entire functions

Denote G = F ∩ H. So, P2 (G) = 1. Then for fixed ω20 ∈ G

P1 (Af (ω20 )):=P1 {ω1 : (∃r0 (ω1 , ω20))(∀r > r0 (ω1 , ω20)) [Nf (r, ω1 , ω20) ≥ ln Sf (r, ω20 ) − (C + 1) ln ln Sf (r, ω20 )]} = 1.

Let P be a direct product of the probability measures P1 and P2 defined on (Ω1 × Ω2 , A1 × A2 , P1 × P2 ). Here A1 × A2 is the σ-algebra, which contains all A1 × A2 such that A1 ∈ A1 and A2 ∈ A2 . By Fybini’s theorem ! ! Z Z Z Z P (Af ) = dP1 (ω1 ) dP2 (ω2 ) ≥ dP1 (ω1 ) dP2 (ω2 ) = Ω2

G

Af (ω2 )

=

Z

Af (ω2 )

dP2 (ω2 ) = P2 (G) = 1.

G

Suppose that there exists a set B1 ∈ Ω1 such that P1 (B1 ) > 0 and for all fixed ω10 ∈ B1 P2 (Af (ω10 )) = P2 {ω2 : (∃r0 (ω10 , ω2 ))(∀r > r0 (ω10 , ω2 )) [Nf (r, ω10 , ω2 ) ≥ ln Sf (r, ω2 ) − (C + 1) ln ln Sf (r, ω2 )]} < 1.

Then P (Af ) =

Z

B1

Z

!

dP2 (ω2 ) dP1 (ω1 ) +

Af (ω1 )


r0 (ω)) [Nh (r, ω) ≥ ln Sh (r, ω)−(C +1) ln ln Sh (r, ω)]} = 1.

Remark that the sequence {ε∗n ηn (ω)} is the sequence of independent random variables and sequences {ε∗n ηn (ω)}, {ηn (ω)} are similar. It remains to denote ζn (ω) = ε∗n ηn (ω).

12

A. O. Kuryliak, O. B. Skaskiv

Theorem 4.6. Let f be a random entire function of the form (3) such that a0 6= 0. Then there exists r0 > 0 such that for all r ∈ (r0 ; +∞) we get p0 (r) ≥ s(r) + N(r) ln N(r) − 4N(r). Proof of Theorem 4.6. By Jensen’s formula we get almost surely 0=

Z

0

r

n(t, ω) 1 dt = t 2π

Z

2π

ln |f (reiθ , ω)|dθ − ln |a0 ξ0 (ω)|, 0 Z 2π 1 ln |a0 ξ0 (ω)| = ln |f (reiθ , ω)|dθ. 2π 0

Therefore, Z 2π o n 1 ln |f (reiθ , ω)|dθ . P {ω : n(r, ω) = 0} ≤ P ω : ln |a0 ξ0 (ω)| = 2π 0 Let G1 = {ω : ln |a0 ξ0 (ω)| ≥ ln γ(ω)}, G2 = o ln γ(ω) , where γ(ω) > 1. Then

n

ω:

1 2π

R 2π 0

ln |f (reiθ , ω)|dθ ≤

Z 2π o 1 iθ G1 G2 = ω : ln |f (re , ω)|dθ > ln γ(ω) > ln |a0 ξ0 (ω)| , 2π 0 Z 2π n o \ 1 G1 G2 ⊂ ω : ln |f (reiθ , ω)|dθ 6= ln |a0 ξ0 (ω)| , 2π 0 Z 2π n o \ [ 1 G1 G2 = G1 G2 ⊃ ω : ln |f (reiθ , ω)|dθ = ln |a0 ξ0 (ω)| . 2π 0 \

n

So, P {ω : n(r, ω) = 0} ≤ P (G1 ∪ G2 ) ≤ P (G1) + P (G2).

(9)

We define Z 2π n 1 ln |f (reiθ , ω)|dθ ≥ A = ω : (∃r0 (ω))(∀r > r0 (ω)) 2π 0 o ≥ ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω) + ln |a0 ξ0 (ω)| . By Corollary 4.5 we obtain that P (A) = 1. Put γ(r, ω) = C1 · |a0 | · |ξ0 (ω)|, C1 > 1. Then we may calculate the probability of the event G1 P (G1 ) = P {ω : ln |a0 ξ0 (ω)| ≥ ln C1 + ln |a0 ξ0 (ω)|} = P {ω : ln C1 ≤ 0} = 0

13

Zeros distribution of gaussian entire functions

and estimate the probability of the event G2 as r → +∞ P (G2 ) = P (G2 ∩ A) + P (G2 ∩ A) ≤ P (G2 ∩ A) + P (A) = P (G2 ∩ A) = n h = P ω : (∃r0 (ω))(∀r > r0 (ω)) ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω)+ Z 2π io 1 + ln |a0 ξ0 (ω)| ≤ ln |f (reiθ , ω)|dθ ≤ ln γ(r, ω) = 2π 0 = P {ω : (∃r0 (ω))(∀r > r0 (ω)) h i ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω) + ln |a0 ξ0 (ω)| ≤ ln C1 + ln |a0 ξ0 (ω)| } = h i = P {ω : (∃r0 (ω))(∀r > r0 (ω)) ln Sf (r, ω) − (C + 1) ln ln Sf (r, ω) ≤ ln C1 } ≤ h i ≤ P {ω : (∃r0 (ω))(∀r > r0 (ω)) ln Sf (r, ω) ≤ 2 ln C1 } = i h = P {ω : (∃r0 (ω))(∀r > r0 (ω)) Sf (r, ω) ≤ C12 } = P {ω : Sf (r, ω) ≤ C12 } ≤ n o X 2 2 2n 4 ≤ P ω: |ξn (ω)| |an | r ≤ C1 . (10) n∈N (r)

The function of distribution of the random variable |ξn (ω)| √ F|ξn | (x) = 1 − exp{−x2 }, F|ξn |2 (x) = F|ξn | ( x) = 1 − exp{−x}, n  x  x o = 1 − exp − F|ξn |2 |an |2 r2n (x) = F|ξn |2 |an |2 r 2n |an |2 r 2n

for n 6∈ N ′ and x ∈ R+ . Then for the random vector η(ω) = (|ξ1 (ω)|a1r j1 , . . . , ξjk (ω)|ajk r jk ), jk ∈ N (r), the density function n o  Q N (r) xn 1  exp − , x ∈ R+ , 2 2n 2 2n |an | r |an | r pη (x) = n∈N (r)  N (r) 0, x 6∈ R+ .

So,

n P ω: =

Y

X

n∈N (r)

n∈N (r)

o |ξn (ω)|2|an |2 r 2n ≤ C14 = P {ω : η(ω) ∈ B(r)} =

1 · |an |2 r 2n

Z

··· B(r)

Z

Y

n∈N (r)

n exp −

xn o dx1 . . . dxk ≤ |an |2 r 2n

≤ exp(−s(r)) · volRN(r) B(r), where B(r) =

(

N (r)

x ∈ R+

:

X

n∈N (r)

(11) )

xn ≤ C14 .

14

A. O. Kuryliak, O. B. Skaskiv

For C > 0 by elementary calculation we get n n o Cn X n volRn x ∈ R+ : xi ≤ C = . n! i=1

From this equality and Stirling’s formula n! =

√

 n n n θ o n , θn ∈ [0, 1], n ∈ N, · exp − 2πn e 12n

it follows that the volume of the set B(r)

  1 1 1 + ln volRN(r) B(r) ≤ − ln(2π) − ln N(r) − N(r) ln N(r) + 2 2 12N(r) +N(r) + 4N(r) ln C1 ≤ −N(r)(ln N(r) − 1 − 4 ln C1 ). Let us choose C1 = 2. From (11) it follows p0 (r) ≥ s(r) + N(r) ln N(r) − 4N(r). Using Lemma 3.3 from Theorems 4.1 and 4.6 we deduce such a statement. Theorem 4.7. Let ε > 0, and f be a random entire function of the form (3) such that a0 6= 0. Then there exist r0 > 0 and the set E ⊂ (1; +∞) of finite logarithmic measure such that for all r ∈ (r0 ; +∞) \ E we obtain p (1 − ε)N(r) ln N(r) ≤ p0 (r) − s(r) ≤ N(r) exp{(2 + ε) ln N(r)},

in particular,

1 ln(p0 (r) − s(r)) ln(p0 (r) − s(r)) , lim ≤ r→+∞ ln s(r) ln s(r) 2 r→+∞

0 ≤ lim

r ∈E /

r ∈E /

and

(12)

ln(p0 (r) − s(r)) = 1. r→+∞ ln N(r) lim

r ∈E /

5

Examples on sharpness of inequalities (12)

Theorem 5.1. There exist a random entire function of form (3) for which a0 6= 0 and a set E of finite logarithmic measure such that ln(p0 (r) − s(r)) 1 = . r→+∞ ln s(r) 2 lim

r ∈E /

15

Zeros distribution of gaussian entire functions

Proof. Consider the entire function f (z) = 1 +

+∞ X zn . n n 2 ) ( n=1 2

For this function and r ∈ (r0 (ω); +∞) \ E we have p p p s(r) 1 ln N(r) = . < N(r) < s(r) exp{(1 + ε) ln s(r)}, lim 3 r→+∞ ln s(r) 2 ln s(r) r ∈E / By Theorem 4.7 we have for r ∈ (r0 ; +∞) \ E p ln N(r) + 3 ln N(r) − ln 2 + ln N(r) + ln ln N(r) ln(p0 (r) − s(r)) ≤ ≤ , ln s(r) ln s(r) ln s(r) ln N(r) 1 ln(p0 (r) − s(r)) = lim = . lim r→+∞ r→+∞ ln s(r) ln s(r) 2 r ∈E /

r ∈E /

Theorem 5.2. There exist a random entire function of form (3) for which a0 6= 0 and a set E of finite logarithmic measure such that ln(p0 (r) − s(r)) = 0. r→+∞ ln s(r) lim

r ∈E /

Proof. Consider the entire functions +∞ X X zn zn f (z) = 1 + n , g(z) = 1 + n , (n) 2 (n)2 n=1 2 n∈N ∗ 2

where N ∗ = {n : n = [ek ] + 1 for some k ∈ Z+ }. Here [ek ] means the integral part of the real number ek . We denote Nf (r) = {n ∈ Z+ : ln(|an |r n ) > 0} \ {0}, Ng (r) = {n ∈ N ∗ : ln(|an |r n ) > 0},  n − n2 X X sf (r) = 2 ln(|an |r n ), sg (r) = 2 ln(|an |r n ), an = , n ∈ N. 2 n∈Nf (r)

n∈Ng (r)

Remark that the sequence {(n/2)−n/2 } is log-concave and Nf (r) = {1, . . . , Nf (r)}. Then by the definition of Ng (r) we get Ng (r) ≤ 2 ln Nf (r), r → +∞. For r ∈ (r0 ; +∞) \ E we obtain (see [22]) Ng (r) ≤ 2 ln Nf (r) ≤ 2 ln(ln µf (r) ln2 (ln µf (r))) < 4 ln ln µf (r).

16

A. O. Kuryliak, O. B. Skaskiv

Remark that min{n ∈ N ′ : n > νg (r)} ≤ [eνg (r)] + 1 < (e + 1) ln νg (r). Let us fix r > 0. Consider the function y(t) = ln(a(t)r t ) = − 2t ln( 2t ) + t ln r, for which a(n) = an . The graph of the function y(t) passes through the points (0; 0) and (νg (r), ln µg (r)). It follows from log-concavity of the function y(t) that the point (νf (r), ln µf (r)) belongs to the triangle with the vertices (νg (r), ln µg (r)), ((e + 1)νg (r), ln µg (r)) and ((e + 1)νg (r), (e + 1) ln µg (r)). Then ln µf (r) ≤ (e + 1) ln µg (r), sg (r) ≥ 2 ln µg (r) ≥

2 ln µf (r). e+1

For the function g(z) and r ∈ (r0 ; +∞) \ E we get ln Ng (r) ln(4 ln ln µf (r)) ln(p0 (r) − sg (r)) = lim ≤ lim = 0. r→+∞ ln sg (r) r→+∞ ln( 2 ln µf (r)) r→+∞ ln sg (r) e+1

0 ≤ lim

r ∈E /

r ∈E /

6

r ∈E /

Zeros distribution of non-gaussian entire functions

One can ask what happens when random variables ξn (ω) in (3) have not gaussian distribution([21])? The following result shows that the situation may be very different. P n Theorem 6.1. Let f (z, ω) = +∞ n=0 ζn (ω)an z , a0 6= 0, with a sequence of independent identically distributed random variables (ζn (ω))n∈Z+ such that 1) (arg ζn (ω))n∈Z+ are uniformly distributed on [−π, π); 2) M|ζn | < +∞ and M( |ζ1n | ) < +∞, n ∈ Z+ ; 3) there exists ε > 0 such that for any n ∈ Z+ we have P {ω : |ζn (ω)| < ε} = 0. Then there exists r0 > 0 such that for all r > r0 we get P {ω : n(f, ω) = 0} = 0. Proof. From inequality (10) we get for some constant C3 > 0 n o X 0 ≤ P {ω : n(f, ω) = 0} ≤ P ω : |ζn (ω)|2|an |2 r 2n ≤ C3 ≤ n∈N (r)

n ≤ P ω:

X

n∈N (r)

o ε2 |an |2 r 2n ≤ C3 = 0 (r → +∞).

Zeros distribution of gaussian entire functions

17

References [1] J.E. Littlewood, A.C. Offord, On the distribution of zeros and a-values of a random integral function, Ann. Math., 49, 50 (1948), 885 - 952, 990 991. [2] J.E. Littlewood, Collected Papers, Oxford, V.2, 1982. [3] A.C. Offord, The distribution of the values of an entire function whose coefficients are independent random variables, Proc. London Math. Soc., 14A (1965), no.3, 199 - 238. [4] A.C. Offord, The distribution of zeros of series whose coefficients are independent random variables, Indian J. Math., 9 (1967), no.1, 175 - 196. [5] A.C. Offord, The distribution of the values of a random functions in the unit disk, Stud. Math., 41 (1972), 71 - 106. [6] A.C. Offord, Lacunary entire functions, Math. Proc. Cambr. Phil. Soc., 114 (1993), 67 - 83. [7] M. Sodin, B. Tsirelson, Random complex zeros. I. Asymptotic normality, Israel J. Math., 144 (2004), 125 - 149. [8] M. Sodin, B. Tsirelson, Random complex zeros. II. Perturbed lattice, Israel J. Math., 152 (2006), 105 - 124. [9] M. Sodin, B. Tsirelson, Random complex zeros. III. Decay of the hole probability, Israel J. Math., 147 (2005), 371 - 379. [10] Yu. Peres, B. Virag, Zeros of i.i.d. gaussian power series: a conformally invariant deteminantal process, Acta Mathematica, 194 (2005), 1 - 35. [11] Yu. Peres, B. Virag, Zeros of i.i.d. gaussian power series and a conformal invariant deteminantal process, arXiv:math.PR/0310297. – 2005. [12] M.P. Mahola, P.V. Filevych, The value distribution of a random entire function, Mat. Stud., 34 (2010), no.2, 120 - 128. [13] M.P. Mahola, P.V. Filevych, The angular value distribution of random analytic functions, Mat. Stud., 37 (2012), no.1, 34 - 51. [14] M.P. Mahola, P.V. Filevych, The angular distribution of zeros of random analytic functions, Ufa. Math. Journ., 4 (2012), no.1, 115 - 127. [15] M. Sodin, Zeros of gaussian analytic functions, Math. Res. Let., 7 (2000), no.4, 371 - 381.

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[16] M. Sodin, Zeros of Gaussian analytic functions, European Congress of Mathematics, Eur. Math. Soc., Zurich, 2005, 445 - 458. [17] M. Sodin, Zeroes of Gaussian analytic functions, arXiv: 0410343v1. – 2004. [18] F. Nazarov, M. Sodin, A. Volberg, Transportation to random zeroes by the gadient flow, Geom. Funct. Anal., 17 (2007), no.3, 887 - 935. [19] F. Nazarov, M. Sodin, A. Volberg, The Jancovici-Lebowitz-Manificat law for large fluctuations of random complex zeroes, Comm. Math. Phys., 284 (2008), no.3, 833 - 865. [20] M. Krishnapur, Overcrowding estimates for zeroes of random analytic function, J. Stat. Phys., 124 (2006), no.6, 1399 - 1423. [21] A. Nishry, The hole probability for gaussian entire functions, arXiv: 0903.4970v1. – 2009. [22] A. Nishry, Asymptotics of the hole probability for zeros of random entire functions, arXiv: 0909.1270v1. – 2009. [23] A. Nishry, Hole probability for entire functions represented by gaussian Tailor series, arXiv: 1105.5734v2. – 2011. [24] A. Nishry, Hole probability for entire functions represented by gaussian taylor series, Journal danalyse mathematique. 118 (2012), 493 - 507. [25] A. Nishry, Topics in the value distribution of random analytic functions, arXiv: 1310.7542v2. – 2013. [26] O.B. Skaskiv, A.O. Kuryliak, The probability of absence zeros in the disc for some random analytic functions, Math. Bull. Shevchenko Sci. Soc., 8 (2011), 335 - 352. [27] A.A. Goldberg, I.V. Ostrovsky, Value distribution of meromorphic functions, Nauka, Moskow, 1970. (in Russian)